Solve left(sqrt{3}-1right){left(sqrt{3}+1right)}^2-sqrt{75}-sqrt{12}-2

$$\left( \sqrt{ 3 } -1 \right) { \left( \sqrt{ 3 } +1 \right) }^{ 2 } - \sqrt{ 75 } - \sqrt{ 12 } -2$$

Evaluate

$$-5\sqrt{3}\approx -8.660254038$$

Solution Steps

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(\sqrt{3}+1\right)^{2}.

$$\left(\sqrt{3}-1\right)\left(\left(\sqrt{3}\right)^{2}+2\sqrt{3}+1\right)-\sqrt{75}-\sqrt{12}-2$$

The square of \sqrt{3} is 3.

$$\left(\sqrt{3}-1\right)\left(3+2\sqrt{3}+1\right)-\sqrt{75}-\sqrt{12}-2$$

Add 3 and 1 to get 4.

$$\left(\sqrt{3}-1\right)\left(4+2\sqrt{3}\right)-\sqrt{75}-\sqrt{12}-2$$

Use the distributive property to multiply \sqrt{3}-1 by 4+2\sqrt{3} and combine like terms.

$$2\sqrt{3}+2\left(\sqrt{3}\right)^{2}-4-\sqrt{75}-\sqrt{12}-2$$

The square of \sqrt{3} is 3.

$$2\sqrt{3}+2\times 3-4-\sqrt{75}-\sqrt{12}-2$$

Multiply 2 and 3 to get 6.

$$2\sqrt{3}+6-4-\sqrt{75}-\sqrt{12}-2$$

Subtract 4 from 6 to get 2.

$$2\sqrt{3}+2-\sqrt{75}-\sqrt{12}-2$$

Factor 75=5^{2}\times 3. Rewrite the square root of the product \sqrt{5^{2}\times 3} as the product of square roots \sqrt{5^{2}}\sqrt{3}. Take the square root of 5^{2}.

$$2\sqrt{3}+2-5\sqrt{3}-\sqrt{12}-2$$

Combine 2\sqrt{3} and -5\sqrt{3} to get -3\sqrt{3}.

$$-3\sqrt{3}+2-\sqrt{12}-2$$

Factor 12=2^{2}\times 3. Rewrite the square root of the product \sqrt{2^{2}\times 3} as the product of square roots \sqrt{2^{2}}\sqrt{3}. Take the square root of 2^{2}.

$$-3\sqrt{3}+2-2\sqrt{3}-2$$

Combine -3\sqrt{3} and -2\sqrt{3} to get -5\sqrt{3}.

$$-5\sqrt{3}+2-2$$

Subtract 2 from 2 to get 0.

$$-5\sqrt{3}$$